Inequalities

Let $f(x) = |x^2-1|+|x^2-4|$ , which is an even function. Therefore, we need only to consider for $x \ge 0$ and reflect the solutions in $x \le 0$ . There are three cases to be considered.

For $0 \le x < 1$ ,

$\begin{aligned} f(x) & = 1-x^2 + 4-x^2 \\ & = 5-2x^2 \\ 5-2x^2 & > 3 \\ \implies x & < 1 \end{aligned}$

For $1 \le x < 2$ , $\begin{aligned} f(x) & = x^2-1 + 4-x^2 = 3 \color{#D61F06}< 3 \end{aligned}$

For $x \ge 2$ ,

$\begin{aligned} f(x) & = x^2-1 + x^2-4 \\ & = 2x^2-5 \\ 2x^2-5 & > 3 \\ \implies x & > 2 \end{aligned}$

Reflecting the solutions in $x \le 0$ , we have $|x^2-1|+|x^2-4| > 3$ , for $x \in (-\infty, -2) \cup (-1,1) \cup (2, \infty)$ . $\implies |a|+|c|+d+e = |-2|+|-1|+1+2 = \boxed{6}$